Question: Let $g(x)=\begin{cases} e^x&\text{for }x \leq -1 \\\\ -e^{-x}&\text{for }x > -1 \end{cases}$ Is $g$ continuous at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Answer: For $g$ to be continuous at $x=-1$, we need $\lim_{x\to -1}g(x)$ and $g(-1)$ to exist and be equal. Since $-1\leq -1$, the rule that applies to $x=-1$ is $e^{x}$. So $g(-1)=e^{-1}=\dfrac{1}{e}$. Now let's analyze $\lim_{x\to -1}g(x)$. Finding $\lim_{x\to -1^{ +}}g(x)$ For $x$ -values larger than $-1$, the appropriate rule for $g(x)$ is $-e^{-x}$. Since $-e^{-x}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -1^{ +}}g(x) \\\\ &=\lim_{x\to -1^{ +}}[-e^{-x}] \gray{-e^{-x}\text{ is the rule for }x>-1} \\\\ &=-e^{-(-1)} \gray{-e^{-x}\text{ is continuous at }x=-1} \\\\ &=-e \end{aligned}$ Finding $\lim_{x\to -1^{ -}}g(x)$ For $x$ -values smaller than $-1$, the appropriate rule for $g(x)$ is $e^x$. Since $e^x$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to -1^{ -}}g(x) \\\\ &=\lim_{x\to -1^{ -}}[e^x] \gray{e^x\text{ is the rule for }x<-1} \\\\ &=e^{-1} \gray{e^x\text{ is continuous at }x=-1} \\\\ &=\dfrac{1}{e} \end{aligned}$ Conclusion We found that $\lim_{x\to -1^{ +}}g(x)=-e$ and $\lim_{x\to -1^{ -}}g(x)=\dfrac{1}{e}$. Since the one-sided limits aren't equal, $\lim_{x\to -1}g(x)$ doesn't exist and $g$ isn't continuous at $x=-1$.